i-ca3d62770cb2e5ec073c2a7ade53f39f-sm_balloon_long_ribbon.jpgAs a follow up to Wednesday’s sad balloon post, the repair that lofted it back to the ceiling was a temporary reprieve, unsurprisingly. After 24 hours, more or less, it had sunk back down to the point where the ribbon was just barely touching the floor.

On the one hand, it looks kind of pathetic again. On the other hand, though, this is a chance to do some more physics. See, I know how much ribbon I clipped off the end to get it to lift off again, and if it’s fallen back to where it’s just touching the floor, that means that the buoyant force due to the displaced air has decreased by an amount equal to the weight of that much ribbon.

So, if we work out how much helium is required to lift that weight, we know how much helium leaked out overnight. Which tells us how quickly the helium inside leaked through the Mylar of the balloon, which is kind of a cool thing to be able to measure with such a simple experiment.

So, how much helium do we need? Well, the net buoyant force is equal to the weight of the displaced air minus the weight of the helium that replaces it. Air has a density of 1.29 kg/m3, while helium has a density of 0.18 kg/m3, so a balloon full of a cubic meter of helium at atmospheric pressure would produce a lift force of:

F=(weight of air)- (weight of helium) = (Volume)((air density)-(He density))(g)

F= (1m3)((1.29 kg/m3)-(0.18 kg/m3))(9.8 N/kg) = 11N

The volume of the balloon is nowhere near a cubic meter, of course, but we can just call this 11 N/m3, and use this to find the volume of helium from the change in lift.

So, how much did the lift change? I could take the ribbon to campus and weigh it on the milligram scale, but that would feel like cheating, so let’s estimate it from the dimensions. I cut 11 in (about 28 cm) of ribbon off to get the balloon to float up to the ceiling again, and the ribbon is about 0.5cm wide. The thickness is too small to measure with the rulers I have here, but I folded it over on itself until there were ten bits of ribbon stacked up, and that stack was about 0.1cm high, so we’ll say that it’s 00.01cm thick (that’s 100 microns, or about the thickness of a human hair, which is probably a little small, but close enough).

If you’re playing along at home, that works out to a total volume of 0.14 cm3, or 0.00000014 m3. To get a mass from this, we need a density. I’m not sure what the ribbon actually it, but it’s sort of similar to paper, so let’s use the density of paper, which the always authoritative Wikipedia gives as 800 kg/m3. Put all that together, and you have an estimated mass of 0.000112 kg, or about a tenth of a gram.

The weight of that length of ribbon, then, is that mass multiplied by the strength of gravity, which is 0.0011 N. Dividing that by our figure of 11 N/m3 for the lifting power of helium suggests that the balloon lost about 9.9×10-5 m3, or about 100 ml.

That seems a little big, but then, it’s not all that much compared to the starting volume of the balloon. When spread out, the balloon has a circular cross-section with a radius of about 20cm, and inflated to about 10cm thick, giving a total volume of 13,000 cm3. If it loses 100 cm3 of helium a day, it would take 130 days for all of the helium to leave (assuming, of course, that the rate is constant.

Now, I probably overestimated the volume of the balloon by considering it as a cylinder. But then, I probably overestimated the density of the ribbon, too, and those two effects will tend to cancel each other out. So this is probably in the right ballpark: a hundred-day lifetime for the complete loss of helium from a Mylar balloon.

That might seem awfully long, but consider this: That particular balloon is left over from my birthday, back in June. So it’s already survived 60 days with at least some of its original helium.

(Of course, this requires the balloon to be treated with respect, or at least benevolent neglect. A similar balloon from SteelyKid’s birthday has already lost much more of its helium, but that was a consequence of an enthusiastic toddler hugging and bouncing it around for a few days.)

So, there’s your latest dose of physics-related overthinking of simple objects…

Comments

  1. #1 Janne
    August 19, 2011

    A related thought experiment, that I – as a non-physicist – would really like an estimate for:

    Assume a vacuum sphere. The inside, containing nothing, has a much lower density than the air it displaces. The spherical container surround it needs to be strong enough to keep it as a vacuum. So, what kind of material would be needed to keep the vacuum intact, but still light enough that you end up with a negative displacement?

  2. #2 becca
    August 19, 2011

    Does the rate of leakage depend on temp and humidity? Empirically, it has seemed so to me.

  3. #3 HP
    August 19, 2011

    When I was a kid, I would tie small weights (usually washers) to the string until the balloon achieved neutral buoyancy. You carefully let go of the balloon, and it Just Stays There. Then I would pretend I was an astronaut aboard Skylab. Good times.

  4. #4 lig tv
    August 19, 2011

    I agree with becca’s comment

  5. #5 anon
    August 20, 2011

    Wouldn’t the loss rate depend on pressure, which would be a function of (mainly) the previous losses? I’d expect to see an exponential decay curve here, like discharging a capacitor through a resistor.

  6. #6 Chad Orzel
    August 20, 2011

    The leakage rate might depend on heat and humidity, but I haven’t tracked that. It’s been inside the whole time, though, and we have central air, so there probably hasn’t been that much variation.

    As for the pressure, unless I’m making a major mistake, I think the pressure inside the balloon is constant at one atmosphere. As the helium inside leaves, the balloon shrinks, keeping the pressure constant at the one atmosphere of the air in the room. If the pressure was higher, the balloon would expand until the combination of air pressure from outside and elastic force from the skin matched the outward force from the gas. If the pressure inside was lower, it would get crushed down to a smaller size by the air pressure, like in that classic lecture demo.

    The partial pressure of helium inside probably drops– I assume there’s at least some ordinary air in there as well– which might affect the rate a little, but probably not all that much until the late stages of the process.

  7. #7 Frank
    August 20, 2011

    Balloons are great for teaching physics to the kids. Every time we have helium balloons that start to sink, my kids make me pull out the hair dryer to heat them up so they’ll float again – the poor man’s hot air balloon. Lasts about 10 seconds.

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